In this paper we show that sufficiently similar finite metric spaces can be amalgamated so that the distance between them is sufficiently small

Ivanov, A.

Majcher-Iwanow, B.

Algebra and Discrete Mathematics

Наукова електронна бібліотека періодичних видань НАН України (Vernadsky National Library of Ukraine)

Algebra and Discrete Mathematics RESEARCH ARTICLEVolume 22 (2016). Number 2, pp. 233–239© Journal “Algebra and Discrete Mathematics”An amalgamation property for metric spacesAleksander Ivanov∗ and Barbara Majcher-Iwanow∗∗Communicated by V. I. SushchanskyAbstract. In this paper we show that suficiently similarfinite metric spaces can be amalgamated so that the distance betweenthem is sufficiently small.1. IntroductionThe following is usually called the free amalgamation property forfinite metric spaces, see Theorem 2.1 of [1].Assume that (X, d1) and (Y, d2) are finite metric spaces with Z = X∩Yand d1 = d2 for elements of Z. Then there is a metric d on X ∪ Y whichagrees with d1 on X, with d2 on Y and is defined for x ∈ X \ Y andy ∈ Y \X byd(x, y) = minz∈Z(d1(x, z) + d2(y, z)) when Z 6= ∅;d(x, y) = d1(x, x∗) + d(x∗, y∗) + d2(y∗, y) when Z = ∅and x∗ ∈ X, y∗ ∈ Y are distinguished togetherwith the distance between them.It has been applied in a number of places. We recommend [4] and [5]for some information concerning applications in the case of the Urysohnspace of diameter 1, U.∗Supported by Polish National Science Centre grant DEC2011/01/B/ST1/01406∗∗Partially supported by ESF Short Visit Grant no. 54192010 MSC: 54A05, 54D80.Key words and phrases: amalgamation, finite metric spaces.234 Amalgamating metric spacesThe main result of our paper is an amalgamation property whichroughly states that ifA andB are finite metric spaces which are sufficientlysimilar then there is a metric on A∪B extending the metrics of A and Bso that A and B are sufficiently close in A ∪B.Some version of this theorem was presented in Section 3 of the firstversion of the preprint [3]. In that paper we apply it in order to evaluatethe distance between types in Th(U) with parameters. It is worth notingthat the parameter-free case (i.e. when A ∩B = ∅) was also consideredin [2], where the construction was based on Example 56 from [6], p. 295–296. Although the general case looks slightly more complicated than thecase from [2], we have found that the idea of Example 56 of [6] worksin the general case too. This improves the corresponding result from [3]and simplifies the proof. A model-theoretic version of this material canbe found in Section 6 of the latest version of [3].Since the essence of this material is elementary we have decided topresent it in this paper in the simplest form. Below we do not use anymathematics beyond elementary metric topology.2. AmalgamationThe following proposition states the property that if A and B arefinite metric spaces which are sufficiently nice and sufficiently similar thenthere is a metric space C containing A and a copy of B under an isometryfixing A ∩B so that A and the image of B are sufficiently close in C.Proposition 1. Let a finite metric space A = {a1, . . . , an} and numbers0 6 q < n and ε > 0 satisfy all inequalities of the form4ε < d(ai, aj) for pairs i < j 6 n with q < j and4ε < d(ai, aj)+d(ai, ak)−d(aj , ak) for triples ai, aj , ak with |{i, j, k}| = 3and k 6 q < min(i, j) < n.Let B be an n-element metric space consisting of elements bi so that foreach pair i < j 6 n, |d(bi, bj)−d(ai, aj)| 6 ε. We assume that a1 = b1,. . . ,aq = bq, A ∩B = {a1, . . . , aq}, and the metric defined on {b1, . . . , bq} inthe space B coincides with the metric defined on {a1, . . . , aq} in A.Then there is a metric on A ∪ B extending metrics in A and B sothat for each q < i 6 n, d(ai, bi) = ε.A. Ivanov, B. Majcher-Iwanow 235Proof. The case A ∩ B = ∅ is already considered in Example 56 [6], p.295 - 296. We will assume that A∩B 6= ∅. Let us build a metric space on{a1, . . . , an, bq+1, . . . , bn} so that d(ai, bi) = ε for all q < i 6 n. We defined(ai, bj) for i 6= j with i > q or j > q as follows:d(ai, bj) = min({d(ai, ak) + d(ak, bj) : k 6 q}∪{d(ai, ak) + ε+ d(bk, bj) : k > q}).Below we will use the observation that in the cases i 6 q < j or j 6 q < ithe metric on B or A respectively which is given in the formulation ofthe proposition, satisfies the condition of this definition. To see this oneshould note that in these casesd(ai, bj) 6 d(ai, ak) + ε+ d(bk, bj) when k > q.Indeed, if for example i 6 q < j thend(ai, bj) 6 d(bi, bk) + d(bk, bj) 6 d(ai, ak) + ε+ d(bk, bj).Let us verify the triangle inequality. We may restrict ourselves by triangleswhich intersect both A \B and B \ A. We will use below the followingconsequence of the assumptions of the proposition:3ε < d(bi, bj) for pairs i < j 6 n with q < j andε < d(bi, bj) + d(bi, bk)− d(bj , bk) for triples bi, bj , bk with |{i, j, k}| = 3and k 6 q < min(i, j) < n.Case 1. d(ai, bj) 6 d(ai, al) + d(al, bj). By the assumptions of theproposition we may assume that i 6= j. If d(al, bj) = d(al, ak) + d(ak, bj)with k 6 q, thend(ai, al) + d(al, bj) = d(ai, al) + d(ak, al) + d(ak, bj) >d(ai, ak) + d(ak, bj) > d(ai, bj).If d(al, bj) = d(al, ak) + ε+ d(bk, bj) for some k > q, thend(ai, al) + d(al, bj) = d(ai, al) + d(ak, al) + ε+ d(bk, bj) >d(ai, ak) + d(bk, bj) + ε > d(ai, bj).Case 2. d(ai, bj) 6 d(ai, bl) + d(bl, bj). This case is similar to Case 1.236 Amalgamating metric spacesCase 3. d(ai, aj) 6 d(ai, bl) + d(aj , bl). We may assume that i > q orj > q. If k,m 6 q andd(ai, bl) = d(ai, ak) + d(ak, bl) and d(aj , bl) = d(aj , am) + d(am, bl)thend(ai, bl) + d(aj , bl) > d(ai, ak) + d(ak, bl) + d(aj , am) + d(am, bl) >> d(ai, ak) + d(ak, am) + d(am, aj) > d(ai, aj).Ifd(ai, bl) = d(ai, ak) + d(ak, bl) with k 6 q andd(aj , bl) = d(aj , am) + ε+ d(bm, bl) with m > q,then j > q < l (otherwise we are in the previous situation) andd(ai, bl) + d(aj , bl) > d(ai, ak) + d(ak, bl) + d(aj , am) + ε+ d(bm, bl) >> d(ai, ak) + d(ak, al)− ε+ d(am, aj) + d(am, al).By the assumptions of the proposition al is not between ak and am andε < d(ak, al) + d(am, al)− d(ak, am).Thusd(ai, ak) + d(ak, al)− ε+ d(am, aj) + d(am, al) >d(ai, ak) + d(ak, am) + d(am, aj) > d(ai, aj).Now assume thatd(ai, bl) = d(ai, ak) + ε+ d(bk, bl) with k > q andd(aj , bl) = d(aj , am) + ε+ d(bm, bl) with m > q.We may assume that q < min(i, j, l) (otherwise we are in one of theprevious situations). Thend(ai, bl) + d(aj , bl) > d(ai, ak) + ε+ d(bk, bl) + d(aj , am) + ε+ d(bm, bl) >> d(ai, ak) + d(ak, al) + d(am, aj) + d(am, al) > d(ai, aj).Case 4. d(bi, bj) 6 d(bi, al) + d(al, bj). This case is similar to Case 3.Note that we can use the inequality ε < d(bk, bl) + d(bm, bl) − d(bk, bm)for k 6 q < min(l,m).A. Ivanov, B. Majcher-Iwanow 237We now formulate our main result.Theorem 1. Let A0 = {a1, . . . , aq} be a finite metric space of size q. Letn > q. Consider spacesA = {a1, . . . , aq, . . . , an} and B = {b1, . . . , bq, . . . , bn},where a1 = b1, . . . , aq = bq. Assumemax{|d(bi, bj)− d(ai, aj)| : 1 6 i < j 6 n} 6 ε.Then there is a metric on A ∪B extending metrics in A and B so thatfor each q < i 6 n, d(ai, bi) 6 18ε.Proof. We start with a procedure which finds a minimal subset A′ ⊂ Acontaining A0 so that the distances between pairs of elements of A′ withat least one of them from A′ \ A0 are > 4ε and A is contained in theneighbourhood of A′ of radius 4ε. Let B′ be the subset of B consisting ofelements with the same numbers as elements of A′ in A. Note that thedistances between pairs of elements of B′ are > 3ε and B is contained inthe neighbourhood of B′ of radius 5ε.Now consider all ai ∈ A′ \A0 which appear in triples ai, aj , ak in A′withd(ai, aj) + d(ai, ak)− d(aj , ak) 6 2ε,where |{i, j, k}| = 3 , ak ∈ A0 , ai, aj 6∈ A0.Firstly find indices of such elements in A and B. Then for each index ijof this set apply free amalgamation of A with the two-element subspace{aij , a′ij} where the distance is rational and satisfies2ε 6 d(aij , a′ij) < 4ε.We repeat this procedure for each element aij of our list. We use Theorem2.1 of [1] (see Introduction above) for every amalgamation. Let Aˆ be theresulting space. Removing ai0 , . . . , aij , . . . from (Aˆ \A) ∪A′ we constructA′′ which already satisfies the assumptions of Proposition 1. Indeed, it isnow obvious that if ai, aj , ak is a triple from A′ as above, then a′i, aj , akfulfill the inequalityd(a′i, aj) + d(a′i, ak)− d(aj , ak) > 4ε.Since d(aj , ak) > max(d(ai, aj), d(ai, ak)) the permutation of a′i and ajdoes not change this property. Moreover it is easy to see that no element238 Amalgamating metric spacesof A′ ∩A′′ plays the role of ai in any triple of A′′ intersecting A0 as above.Such an assumption whould imply the false statement that this elementhas the same property in A′ (possibly replacing some a′ij in the triple byaij ). We use here the fact that the space Aˆ satisfies the condition thatany non-trivial path from a′ij contains aij .We apply the same procedure to B′ where we putd(bij , b′ij) = d(aij , a′ij).As a result we obtain the corresponding Bˆ and B′′.We enumerate A′ andB′ according the enumerations ofA andB. Thenwe consider A′′ under the enumeration induced by A′ where the numberof every a′ij is just ij . Note that in Aˆ the distance between elements of A′and A′′ with the same numbers (i.e for example d(aij , a′ij)) is not greaterthan 4ε. In particular A is contained in the neighbourhood of A′′ of radius8ε. Thus in Bˆ the space B is contained in the neighbourhood of B′′ ofradius 9ε. Moreover for each ai ∈ A there are a′ and b′ which have thesame number in A′′ and B′′ respectively such that d(ai, a′) 6 8ε andd(bi, b′) 6 9ε.On the other hand it is easy to see that we still havemax{|d(bi, bj)− d(ai, aj)| : 1 6 i < j 6 n} 6 ε,for the corresponding elements of A′′ and B′′. As a result we have A′′ andB′′ which satisfy the assumptions of Proposition 1. Applying Proposition 1to A′′ and B′′ over A0 we obtain an amalgamated metric on A′′ ∪ B′′,so that the distances between the corresponding elements with numbers> q is ε. Using the free amalgamation property we amalgamate Aˆ withA′′∪B′′ and then amalgamate the result with Bˆ. In this space A is distantfrom A′′ by 6 8ε and B is distant from B′′ by 6 9ε. The rest is clear.References[1] S.A. Bogatyi, Metrically homogeneous spaces, Russian Math. Surveys, 57 (2002),221–240.[2] S. Coskey, M. Lupini, A Lópes-Escobar theorem for metric structures, and thetopological Vaught conjecture, Fund. Math., 234 (2016), no. 1, 55–72.[3] A. Ivanov, B. Majcher-Iwanow, Polish G-spaces and continuous logic. arxiv:1304.5135.[4] J. Melleray, Some geometric and dynamical properties of the Urysohn space, Topol-ogy Appl., 155 (2008), 1531–1560.[5] J. Nesˇetrˇil, Metric spaces are Ramsey, European J. Combin., 28 (2007), no. 1,457–468.[6] P. Petersen, Riemannian Geometry, Graduate Texts in Mathematics, 171, Springer,New York, 2006.A. Ivanov, B. Majcher-Iwanow 239Contact informationA. Ivanov Institute of Mathematics, University of Wrocław,pl.Grunwaldzki 2/4, 50-384 Wrocław, PolandE-Mail(s): ivanov@math.uni.wroc.plWeb-page(s): www.math.uni.wroc.pl/∼ivanov/B. Majcher-Iwanow Institute of Mathematics, University of Opole,ul.Oleska 48, 45-052 Opole, PolandE-Mail(s): bmajcher@math.uni.opole.plReceived by the editors: 17.11.2015and in final form 09.02.2016.

An amalgamation property for metric spaces

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An amalgamation property for metric spaces

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